sort all elements :nlogn
If the last 2 arrays are B and C, then sort elements of the form
(bi+cj):O(n^2) time O(n^2) space
[for the above step, smallest element  in  b1+c1,next element is smaller of
(b1+c2) and(c1+b2), increase pointer accordingly.  If at one step, the
element under consideration is bi and cj, then the next smallest element is
min(bi+c(j+1),b(i+1)+cj) ]
for each element in A, see if corresponding element is there in [B+C]
O(nlogn)
If yes, then search for  the corresponding two elements in array B and C
O(nlogn)

total time O(n^2), space = O (n^2)

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